? Haaaaaaanka: x5−x4−5x3+5x2+6x−6=0

Haaaaaaanka: PS. żle zapisałam ma byc: x⁵−x⁴−5³+5²+6x−6=o

Julek: x⁵−x⁴−5x³+5x²+6x−6=0 x⁴(x−1) −5x²(x−1)+6(x−1) = 0 (x−1)(x⁴−5x²+6) = 0 t = x², t≥0 t²−5t+6 = 0 Δ = 25 − 24 = 1²

	5+1
t1 =		= 3
	2

	5−1
t2 =		= 2
	2

x² = 2 x = −√2 ⋀ x = √2 x² = 3 x = −√3 ⋀ x = √3 Odpowiedź : x∊{−√3;−√2;1;√2;√3}

Eta: x⁴( x−1) −5x²( x −1) +6( x −1)=0 ( x−1)( x⁴ −5x²+6)=0 (x−1)( x⁴ −2x² −3x²+6)=0 ( x−1)[ x²( x²−2) −3( x²−2)]=0 (x−1)( x²−2)(x²−3)=0 (x−1)(x−√2)(x+√2)(x−√3)(x+√3)=0 rozwiązaniami są: x =1 v x = √2 v x = − √2 v x = √3 v x =−√3

Eta:

Haaaaaaanka: dziekuje